Given the function y = \frac{x^2 + 2x}{(3x - 1)^2}, what is the value of \frac{dy}{dx} at … — JAMB UTME

Given the function $y = \frac{x^2 + 2x}{(3x - 1)^2}$ , what is the value of $\frac{dy}{dx}$ at $x = 1$ ?

-\frac{5}{4}

Applying the quotient rule

\frac{dy}{dx} = \frac{v u' - u v'}{v^2}

with

u = x^2 + 2x

(

u' = 2x + 2

) and

v = (3x - 1)^2

(

v' = 6(3x - 1)

) yields

\frac{dy}{dx} = \frac{16 - 36}{16} = -\frac{5}{4}

x = 1

Identify the numerator and denominator functions for the quotient rule

y = \frac{u}{v}

u = x^2 + 2x

and

v = (3x - 1)^2

The function is structured as a quotient of two algebraic expressions.

Differentiate

u

and

v

with respect to

x

\frac{du}{dx} = 2x + 2

and

\frac{dv}{dx} = 2(3x - 1) \cdot 3 = 6(3x - 1)

Use the power rule for

u

and the chain rule for

v

Apply the quotient rule formula

\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}

\frac{dy}{dx} = \frac{(3x - 1)^2 (2x + 2) - (x^2 + 2x) \cdot 6(3x - 1)}{(3x - 1)^4}

The quotient rule formula combines the expressions and their derivatives.

Substitute

x = 1

into the derivative expression and simplify.

\frac{dy}{dx} = \frac{(2)^2 (4) - (3) \cdot 6(2)}{(2)^4} = \frac{16 - 36}{16} = -\frac{20}{16} = -\frac{5}{4}

Evaluating at

x = 1

yields the numerical derivative value.

Quotient Rule and Chain Rule of Differentiation

Question